1.Explain (in your own words) seven good reasons for freezing cell l.pdf

•

0 likes•2 views

1.Explain (in your own words) seven good reasons for freezing cell lines. 2.What are two hazards of liquid nitrogen? Describe which personal protective equipment (PPE) to use and why when working with liquid nitrogen. 3.Explain four ways we can mitigate the harmful effects of intracellular ice crystal formation during cryopreservation. 4.Name one hydrophilic cryoprotectant used in cell culture. 5.What should you include when labeling your cryovial? Solution Answer = cell cultures are preserved by freezing cell lines good reasons 1] cell cultures provide a good host system for propagation of majority of clinically important animal viruses long term preservation is storage under liquid N2 at -196 0C 2] It reduce the risk of microbial contamination 3] for reduction of risk of cross contamination of other cell lines 4] REduction of risk of genetic drift & morphological changes 5] REduced cost 6] work conducted using cells at a consistent passage number 7] For maintaining the viability of culture cells.

Education

Similar to 1.Explain (in your own words) seven good reasons for freezing cell l.pdf

What is Cryopreservation ? Cryopreservation is a process where biological materials such as cells and tissues are preserved by cooling to very low temperatures, usually at -196°C (the temperature of liquid nitrogen), yet remain viable after later warming to temperatures above 0°C. Cryopreservation in aquatic species goes back 65 years and began about the same time as similar research was performed in livestock (Blaxter 2011). In India, NBFGR & CIFA are the primary organization carrying out fish sperm cryopreservation for long term gene banking (J. K. Jena 2012)

Cryopreservation and its application to aquaculture.pptx

Narsingh Kashyap

Aa313 fibra cel_update

Duc Nguyen Manh

Cryopreservation, germplasm storage

KAUSHAL SAHU

Cryopreservation ( methods and application)

jayalakshmi311

Pure culture preservation technique culture preservation cryopreservation lyophilization media transfer subcuture microbiology biotechology pure culture preservation lab skills lyophilization storing at low temperatre storage maintanence culture culture viability culture oppourtunities culture methods storage purpose storade methods microbiology biotechnology biochemistry zoology botany cultivation aerobic anaerobic

preservation.pptx

DhanaLakshmi787194

Immunohistochemistry principle

James Waita

VACCINE FINAL (1).pptx

ArnabPanja4

pub-thebigfreeze-IPT-sept15

biocision

Preservation of industrial cell culture

Rajesh Sagalgile

eye bank, eye donation, corneal transplantation.pdf

AnjuAnnMani1

2019 bt 204 module 1(part d)

Leon Ittiachen

Copyright © 2013 Pearson Education, Inc.Lectures prepared .docx

vanesaburnand

Cryopreservation and reconstitution of preserved cell lines

Abdulrahman Muhammad

Cryopreservation

surehuasb

Skin and oral mucosal substitutes

Kptaiping Perak

CONSERVATION OF MEDICINAL PLANTS

Dr K SUDHEER KUMAR KANDIBANDA

Introduction to Animal Cell Culture

Shailendra shera

Cell lines

Heena Parveen

Eye banking

Burdwan Medical College and Hospital

Immunohistochemistry (IHC) Principle

Boster Biological Technology

Similar to 1.Explain (in your own words) seven good reasons for freezing cell l.pdf (20)

Cryopreservation and its application to aquaculture.pptx

Aa313 fibra cel_update

Cryopreservation, germplasm storage

Cryopreservation ( methods and application)

preservation.pptx

Immunohistochemistry principle

VACCINE FINAL (1).pptx

pub-thebigfreeze-IPT-sept15

Preservation of industrial cell culture

eye bank, eye donation, corneal transplantation.pdf

2019 bt 204 module 1(part d)

Cryopreservation and reconstitution of preserved cell lines

Cryopreservation

Skin and oral mucosal substitutes

CONSERVATION OF MEDICINAL PLANTS

Introduction to Animal Cell Culture

Cell lines

Eye banking

Immunohistochemistry (IHC) Principle

More from mumnesh

Poisson distribution] The number of data packets arriving in 1 sec at.pdf

mumnesh

Problem 1: Create Node class (or use what you have done in Lab4) • Add methods to get and set, Data and Link • Here the data is an object. (java object) your Linked List can hold any java class. a) Create MyLinkedList2 class and write methods to • Methods to insert nodes at the beginning, at the end and at a specific position. insertFirst(Node x), insertLast(Node x) , insertLast(Node x, int pos) • Methods to delete node from the beginning and from the end. delFirst(), delLast() • Method to search the Linked List for a specific object. searchList(x Object) • Method to Sort the Linked List in ascendant or descendant order according to the object value(s). for example sorted by name, by age, by gpa. sortAsc(),sortDesc(). • Method to display all nodes objects (printing an object is related to its toString() method). • Method to return the largest and smallest object in the list – maxList(), minList() according to some specific object members. . • Method to return the number of nodes – listSize() • Method to merge two linked Lists into one. MyLinkedList2 list3= mergeList(MyLinkedList2 list1, MyLinkedList2 list2) b) Create Main class : MyLinkedList2Demo • Create a linked List LL1 and insert some objects to it. Use any a student java class. • Fill LL1 by adding students from an input file. • Display all the objects (students) in the List LL1 • Search for a particular object (Student by name or by Id or both) in LL1 and print true or false • Sort LL1 and display it ( using asc and desc on multiple members) • Perform all the delete methods on LL1. • Perform all the insert methods on LL1. • Call and run all the implemented methods in a sequential order. NB: Organize a nice output for all the questions above. Problem 2: Same as Pb1 using double linked lists. Create Nod3e and MyLinkedList3 classes. -------------------------------------------------------------------------------------------------------------------- --------------------------------------------------- The input file : id fname lname age gpa year college cuCourses totalCourses 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201112345 Amna chamisi 17 2.1 2014 ENG 31 32 201412345 Amna chamisi 17 2.1 2014 COE 5 29 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 Solution /** * Node.java * @author * */ public class Node { public Object getObj() { return obj; } public Node setObj(Object obj) { this.obj = obj; return this; } public Node getNext() { return next; } public Node setNext(Node next) { this.next = next; return this; } private Object obj; private Node next; } import java.util.Collections; import java.util.Comparator; import java.util.LinkedList; /** * MyLinkedList2.java * @author * */ public class MyLinkedList2 { public void insertFirst(Node x) { if (head == null) { head = x; } else { x.setNext(head); } } public void insertLast(Node x) { if (head == null) { head = x; r.

Problem 1 Create Node class (or use what you have done in Lab4)• .pdf

mumnesh

One of your friends has several brothers and sisters, each quite different in appearance, interests, and abilities, even though all have the same parents. Explain in detail how these siblings reflect crossing over and independent assortment. Solution Meiosis process increases the genetic diversity in a species. In prophase I, crossing over takes place and in metaphase I and anaphase I, independent assortment occurs. When homologous chromosomes are paired together, during meiosis, the chromosomes make contact with the other pair of chromosomes. This point of contact is responsible for the exchange of genetic information between the chromosomes and is called the chiasmata. Although the same genetic information is contained by the two homologous chromosomes, the assortment of the chromosomes determine which information is present in each of the four gametes produced. For a particular phenotype, an allele can determine which characteristic will be expressed by the organism. For example, Chromosome 1 contains an allele for brown eyes while chromosome 2 contains an allele for black eyes. Thus, crossing over and independent assortment are responsible for different appearence, abilities and interest..

One of your friends has several brothers and sisters, each quite dif.pdf

mumnesh

Malpighiaceae luonanthaceae Ochnaceae Flacourtiaceae O Monilophytes Gymnosperms O Basal Angiosperms Monocots Basal Eudicots Asterids Rosids All Plots Escaloniacea Pittosporaceae Araliaceae Caprifoliaceae Adoxaceae Garry aceae Rubiaceae Apocynaceae Solution There are 7 groups present in the legend around the phylogeny.Monophyletic taxon is one that include a group of organism desecended from a single ancestor.A monophyletic taxon is called a clade.Monophyletic groups are important to understand how animals are classified. It is commonly known that all animals evolved over time. The monophyletics in the legends are Cyatheaceae Commonly known as the scaly tree ferns,Poaceae,Arecaceae are a botanical family of perennial climbers, shrubs, acaules and trees commonly known as palm tree,Liliaceae are mostly perennial herbs from starchy rhizomes, corms, or bulbs comprising about 280 genera and 4,000 species. earthworms, birds, and snakes all have a common ancestor. Since all these organisms share a common ancestor, they also share a synapomorphy, which is that earthworms, birds, and snakes all reproduce sexually. These are all come under phylogeny of monophyletic..

Malpighiaceae luonanthaceae Ochnaceae Flacourtiaceae O Monilophytes G.pdf

mumnesh

In Java: Write a program that reads a text file that contains a grade (for instance, 87) on each line. Calculate and print the average of the grades. Solution import java.io.BufferedReader; import java.io.FileReader; // importing filereader import java.io.IOException; // importing file io exceptions public class grade { public static void main(String[] args) { BufferedReader br = null; double count=0; int noOfgrades=0; try { String sLine; br = new BufferedReader(new FileReader(\"sample.txt\")); // reading file using buffer reader while ((sLine = br.readLine()) != null) { // rading line by line noOfgrades=noOfgrades+1; // getting number of grades int num=Integer.parseInt(sLine); // convert string into integer count=count+num; // adding each grade to count } System.out.println(\"count:\\t\"+count); System.out.println(\"noOfgrades:\\t\"+noOfgrades); System.out.println(\"Average of the grades:\\t\"+(count/noOfgrades)); } catch (IOException e) { e.printStackTrace(); // catch exception } finally { try { if (br != null)br.close(); // closing buffer } catch (IOException ex) { ex.printStackTrace();// printing exceptions } } } } /********** out file****** fileName:sample.txt file contains \" 23 34 56 7 88 89 52 34 65 \" */ /************output************* count: 448.0 noOfgrades: 9 Average of the grades: 49.77 */.

In JavaWrite a program that reads a text file that contains a gra.pdf

mumnesh

How does Pb and TCE partition between components of soil (mineral, organic matter, air, and water fractions)? Where can each pollutant can be found if it is spilled on to a loam soil with about a neutral ph, and about 2% organic matter content? Solution lead is heavy metal as a mineral in soil while trichloroethylene is an ogranic pollutant present as organic matter lead will be found in as a mineral into the soil while TCE is present in air as organic pollutant..

How does Pb and TCE partition between components of soil (mineral, o.pdf

mumnesh

Hi, please I need help with the correct answer to the above multiple questions. Thanks. Some animals, such as snakes, can see infrared waves. The un human eye can not detect these waves, however, we can aided detect some wavelengths of light. Considering the following statements, which, if any, would be true? (SELECT ALL TRUE STATEMENTS) O if an object is seen as red, red light is absorbed by the object and not reflected O Visible light waves exist in the 100- 300nm wavelength range C ight is a form of electromagnetic energy that slows down as it enters a medium that is less dense O if an object is green, the object is absorbing non-green waves and reflecting green waves. since humans can not see all wavelengths of the electromagnetic spectrum, only waves within the visible light spectrum will damage the eye e. UV light can not damage the retina because we can not see UV light Solution 1. If an object is green, the object is absorbing non-green waves and reflecting green waves. The reflected green waves are captured by our eyes and the object appears \'green\'. The first statement is the exact opposite of this and, hence, incorrect. 100-300 nm wavelengths correspond to the UV spectrum. Light speeds up in a rarer medium. And, UV light can (seriously) damage our eyes. 2. Superior colliculus: This part of the brain is responsible for saccades - the process by which our eyes scan different images of an object with different focus and \'knits\' them together to perceive useful information. 3. Lens cannot be transplanted! All other options are invalid. 4. Tympanic membrane: Commonly known as the \'ear drum\', this thin membranous structure is present at the end of the external ear canal. All other structures listed are present beyond the tympanic membrane. When something such as a Q-tip or earbud is inserted too deep in the ear canal, it will first perforate the tympanic membrane before injuring any other part..

Hi, please I need help with the correct answer to the above multiple.pdf

mumnesh

For which phyla are megaspores a shared trait? Bryophyta Monilophyta Coniferophyta Anthophyta For which phyla are microspores a shared trait? Bryophyta Monilophyta Coniferophyta Anthophyta For which phyla are sporophyte a shared trait? Bryophyta Monilophyta Coniferophyta Q Anthophyta Solution 13.Bryophta (Bryophytes do not have flowers or seeds and reproduce through spores.) 14.Coniferophyta( A pine tree produces two types of spores- the microspores seen on male cones and megaspores seen on female cones. Hence, they are heterosporic.). 15.Bryphyta(The sporophyte dries and releases spores that grow into a new generation of gametophytes. The sporophyte is the organ for asexual reproduction and consists of a foot, seta and capsule. The capsule is a beak like structure found on top of a thin stalk and produces the spores. Mosses show alternation of generation between gametophytic and sporophytic generations though the plant has a gametophyte dominant cycle.).

For which phyla are megaspores a shared trait Bryophyta Monilophyt.pdf

mumnesh

For Programming Embedded Systems Q-07 *Whcih variable and condition is sued to detect unfinished tasks (interrupted interrupt)? A) Line 28 B) Line 36 C) Line 25 D) Line 22 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I Transitions switch (state) case BL Ledoff: 11. task tasks [2] break; case BL Ledon: 13. const unsigned char tasksNum 2; 14. const unsigned long periodBlinkLed = 1500; 15. const unsigned long periodThreeLeds 500; break; default: break; ) // State actions 17, const unsigned long tasksPeriodGCD = 500; return state 19. int TickFct BlinkLed (int state) 20. int TickFct ThreeLeds (int state) 22. unsigned char process!ngRdyTasks = 0; 23. void TimerISR 98. enum TL States TL TO, TL T1, TL T2 TL State; 100 99. int TickFct ThreeLeds (int state) ( 101 VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks) /*e.g., static int x = 0; */ /*variables for state machine go here. printf (\"Period too short \ \"): 103 switch (state) case -1: process!ngRdyTasks = 1; for (1 = 0; 1 = 107 108 109 tasks [i].period) case TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0 break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0; state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1; tasks [1].period = periodBlinkLed; tasks [1] .elapsedTime tasks [1] ·period; tasks [1].TickFct 6T1ckFct BlinkLed; 118 119 120 121 122 123 1 state = TL T0; break; default: state --1; tasks [1] , state = -1; tasks [1].period periodThreeLeas; tasks [1].elapsedTime = tasks [1].period; tasks [1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137 139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1 Solution c) Line 25 is correct Line 25 contains if condition and processingRdyTasks variable is used to detect the unfinished tasks.

For Programming Embedded SystemsQ-07 Whcih variable and condition.pdf

mumnesh

For analysis of Thermodynamic Cycle, we draw T-s or P-h diagram. Why don’t we draw T-h or P-t diagram? Solution For analysis of thermodynamics cycle we draw various diagrams, In all these diagram T-s,P-h,T- h and P-T diagram are some of them. Every diagram give some specific informations about system working like T-s diagram give information about Heat transfer, P-h diagram use in air conditioning because air is mixture so pressure affect more and enthalphy is measure of energy. like these two diagram we use P-t diagram use in pure substance to see the transformation at triple point. and T-h diagram use for visualise how energy change with temprature..

For analysis of Thermodynamic Cycle, we draw T-s or P-h diagram. Why.pdf

mumnesh

Fill in the blank. A contingency table relates only continuous random variables the difference in the means of two random variables two categories of data a particular response with order in which that response should be applied Solution A contingency table relates Two categories of data Contingency table (cross tabulation or crosstab) is a type of table format that displays the frequency distribution of the two or more categorical variables..

Fill in the blank. A contingency table relates only continuous rando.pdf

mumnesh

Discuss the dangers associated with hydrogen sulfide in the petro.pdf

mumnesh

Complete the following paragraph to describe the various reproductive strategies seen in the phylum Ascomycota. Some choices may be used more than once. Solution 1. The Ascomycota contains two main groups; Asexual sac fungi which include bakers yeast and Sexual sac fungi, which include candida yeast and molds. 2. Most Ascomyta are conidospores and reproduced by meiosis, as seen in candida yeast, or through the production of fruiting body as seen in molds. 3. In contrast in haploid a(n) diploid containing two nuclei develops within a(n) Conidiospores . 4. The nuclei fuse creating haploid cell that undergose mitosis to produce diploid cells. 5. Typically at this point, the ascospores swells and burst, releasing the ascus in to the air to germinate once landed..

Complete the following paragraph to describe the various reproductive.pdf

mumnesh

Communications: Systems have four basic properties (holism, equi-finality, negative entropy). List the principle behind each property. What does it mean that “a system is more than the sum of its parts?” How does this concept apply to organizations? Solution Four basic properties of systems are: Principle of Holism: It emphasizes that system is more than sum of its part due to interdependence upon each other. Principle of Equi-finality: It states that outcome can be achieved through many different paths due to interdependence of components. Principle of Negative Entropy: It is based upon openness of the system that says that system owns the ability to prevent deterioration and thrive. Principle of Requisite Variety: It is also based upon openness of the system and it says that internal complexity should be maintained to deal with external complexity. Statement of “a system is more than the sum of its parts” means that system achieves more when all of its parts complement each other and work together. Each component / part of the system cannot do much in isolation but they can achieve big goals when grouped together as a team. It is very important in organizational perspective as it consists of many different functions as components. When, these functions work together, organizational objective is achieved. If not, organization falls apart..

CommunicationsSystems have four basic properties (holism, equi-fi.pdf

mumnesh

Based on your reading of the GCU introduction and the textbooks, what is the main difference between the modern western worldview and the postmodern worldview? Which parts, if any, do you identify with? Explain. Cite references from your reading to support your answer. Solution Postmodernism is a response against modernism. According to everyone has a point of view through which they see the world. Premodern (the worldview before modernism)- says that natural events are caused by supernatural causes. God, angels, demons, and the like are a reality to these folks..

Based on your reading of the GCU introduction and the textbooks, wha.pdf

mumnesh

Answer using Giardias, Shiga-toxin producing E.coli 1.Causative Agnet of Disease : 2.Type of Agent ( eg. Virus,bacterium) : 3.Further Classification ( eg.Gram type,viral serotype) : 4.How is disease diagnosed? 5.How is the disease treated? 6.How can the disease be prevented? 7.Distinguishing characteristics of organism (what makes the organism unique such as virulence factors that are required for disease, unique attributes that aid or prohibit the identificaiton of the organism ): Solution 1). Giardiasis is an infectious disease caused by a parasite called Giardia lamblia. It is spread through eating contaminated food, contact with the infected people, contaminated water and through pets. This organism is generally found thought the worlds, particularly in the developing countries..

Answer using Giardias, Shiga-toxin producing E.coli1.Causative Agn.pdf

mumnesh

Abraham Shine recently opaned his own accounting fiem on April 1, which he operates as a corporation. The name of the new entity is Abraham Shine CPA Shine xparienced the following events during the organizing phase of the new business and ies Srst month of operations in 2018 (Cick the icon to view the transactions) Read the teaments Requirement 1. Analyze the effects of the events on the accounšing equation of Abraham Shine, CPA Anal ze the events chronolo caly one transaction at a rme egning with the transaction on the 6th caklate the balance in each account after analyzing the efect of the transaction on the account g equation Aher calculating the ending balance of each account on the 30m caloulate total assets and totall Babilities and equity (Complete only the nacessary answer boxes for your transaction ines Do not enter any zeros for your transaction lines 1 Car down all balances to the al line including zero balance accounts entering \"Q for any zero balances Enter decrease in ar account with 4 minus sign or pare theses Abbreviations used Ap Accounts Payable A Accounts Receivable. Com Common Contr “Conttuted. Div . Dividends. Exp . Expense. Fum . Fumitu, Liab. Liablies Rev-Revenue Sup . Supplies, Uni-Uilties) ASSETS LIAB+ EQUITY Contr. Retained Earnings ServiceRent Exp Cap Cash + AR + Ofice + Furn AP + Com Die UtI Sup Stock Rev Exp Enter any inumber in the edit fhelds and then click Check Answer 13-taring Clear Al Check Answer Solution Answer Assets = Liabilities + Equity Contributed Capital + Retained earnings Date Cash + A/R + Office Sup + Furn = A/P + Com Stock - Div + Service Revenue - Rent Exp - Util Exp Apr-05 80000 + + + = + 80000 - + - - Apr-06 -200 + + 200 + = + - + - - Apr-07 + + + 7500 = 7500 + - + - - Apr-10 3500 + + + = + - + 3500 - - Apr-11 -390 + + + = + - + - - 390 Apr-12 + 12000 + + = + - + 12000 - - Apr-18 -1600 + + + = + - + - 1600 - Apr-25 12000 + -12000 + + = + - + - - Apr-27 -7500 + + + = -7500 + - + - - Apr-30 -4000 + + + = + - 4000 + - - TOTAL 81810 0 200 7500 0 80000 4000 15500 1600 390 Income Service Revenue 15500 Expenses Rent Expense 1600 Utilities expense 390 Total expense 1990 Net Income 13510 Opening Bal 0 Add: Net Income 13510 Less: Dividend paid 4000 9510 Closing balance 9510 A. Owner\'s Equity Common Stock 80000 Retained Earnings 9510 89510 B. Liabilities Accounts payable 0 Total Owners\' Equity and Liabilities 89510 C. Assets Cash 81810 Accounts Receivables 0 Office Supplies 200 Furniture 7500 Total Assets 89510 Assets = Liabilities + Equity Contributed Capital + Retained earnings Date Cash + A/R + Office Sup + Furn = A/P + Com Stock - Div + Service Revenue - Rent Exp - Util Exp Apr-05 80000 + + + = + 80000 - + - - Apr-06 -200 + + 200 + = + - + - - Apr-07 + + + 7500 = 7500 + - + - - Apr-10 3500 + + + = + - + 3500 - - Apr-11 -390 + + + = + - + - - 390 Apr-12 + 12000 + + = + - + 12000 - - Apr-18 -1600 + + + = + - + - 1600 - Apr-25 12000 + -12000 + + = + - + - - Apr-27 -7500 + + + = -7500 + - + - - Apr-3.

Abraham Shine recently opaned his own accounting fiem on April 1, whi.pdf

mumnesh

29. This figure shows the principal coronary blood vessels. Which one is the left coronary artery (LCA)? B. 8 C. 5 D. 3 E. 1 will cause a more severe myocardial infarction (MI) 30. Obstruction of the than the obstruction of any of the others A. left marginal vein B. left coronary artery (LCA) C. posterior interventricular vein D. anterior interventricular branch E. circumflex branch Solution 29. D. 3 30. B. left coronary artery 31. A. cardiac muscle fibers have striations 32. C. Sinatrial (SA) node. 33. B. . The tendinous cords. 34. e. They have about the same endurance as skeletal muscle fibers. 35. SA node, AV node, AV bundle, Purkinje fibers, cardiocyte in LV 36. A. Na+ inflow 37. E. slow CA2+ channels.

29. This figure shows the principal coronary blood vessels. Which one.pdf

mumnesh

14. Suppose I have collected the names of people in several separate files based on their starting letter. The names in the first file begin with A through G, the names in the second file begin with H through N, the names in the third file beginning with O through T, and the names in the fourth file contain names beginning with U through Z. None of the files are sorted. instead, the names that appear in each file occur in any order. Describe in detail how I could take advantage of multithreading to create one sorted list of all the names. Make sure you include details such as what your Thread class would do and what your main class that uses the Threads would do. (3 points) Solution We can slice strings with negative indexes as well. This IDLE transcript should clarify the rules for slicing where both indexes are specified: >>> word = \"PYTHONISCOOL\" >>> print(word[2:7]) THONI >>> print(word[6:6]) >>> print(word[4:2]) >>> print(word[5:11]) NISCOO >>> print(word[8:12]) COOL >>> >>> print(word[-7:-2]) NISCO >>> print(word[-9:8]) HONIS >>> print(word[-3:-7]) >>> Notice if you attempt to slice a string where the starting index refers to a position that occurs at or after the ending index, the slice is the empty string, containing no characters. A string can also be sliced using only one index. If only one index is given, Python must know if it’s the start or end of the slice. It assumes that the omitted index must refer to the beginning or end of the string, accordingly. These examples should clarify slicing using only one index: >>> print(word[8:]) COOL >>> print(word[-6:]) ISCOOL >>> print(word[:6]) PYTHON >>> print(word[:-4]) PYTHONIS.

14. Suppose I have collected the names of people in several separate.pdf

mumnesh

4. The size of instructions can be fixed or variable. What are advant.pdf

mumnesh

More from mumnesh (20)